The Math Behind Rail Guns

Blog Post Number 10 Written: 05-24-2021 Uploaded:03-28-2022 (Sorry this wasn’t uploaded on 03-25-2022 Like it should have been, life happens.)

I know I said I am bad at math, but that doesn’t stop me from trying. I’d rather try, be wrong, and get corrected than have never tried at all and been even more wrong. So… It might not be relevant in the first book, or even the second book…. But there are rail guns, and here’s a little math I did to try and make sure I described the kinetic weapon’s effects appropriately.

To start with, I’ll talk a little about normal ballistics. A typical projectile’s kinetic energy is not difficult to calculate. For example a simple kinetic energy formula for a bullet fired horizontally in the atmosphere is The weight of the bullet (in grains) multiplied by the velocity squared (in feet per second) and that number is then divided by the number 450440 (this 450440 number is a constant that takes into account average air pressure, temperature and humidity at just above sea level.) Let’s see if Microsoft word will let me write formulas.

W x V^2/450440=ft-lbs

Not everyone is familiar with a ‘grain’ as a unit of measurement, but at least in the manufacture of ammunition that is the standard unit of measure. So for the purposes of this paper, don’t mistake G for grains and G for grams. They are two different units of measurement. There are roughly 7000 grains to a pound and only 453 grams per pound.

As an example, the 30-30 your fudd grandpa shot bambi with should look like this…

170 G x 2200 fps^2 /450440=1826.65 ft-lbs 

Make sense? Tracking so far?

The problem is, that formula doesn’t work with excessive numbers…

My rail gun, on the other hand, does involve some excessive numbers. To start with the projectile must be a Ferris metal in order to be magnetically accelerated, a very hard metal that resists deformation in this case I selected to use Tungsten.

I knew velocity was going to be very high, in order to strike targets as accurately as possible over the vast interplanetary distances involved in space combat, you would want a fast projectile so that flight time would be as short as possible and there would be less time for your target to avoid the shot. I chose a small projectile ( as far as cannons are concerned anyway.) 2 kilograms of tungsten occupies a volume of 103.9 cubic centimeters. However, I am an American science fiction author and an Aviation mechanic, airplanes and the old cars I own, don’t come in metric, I don’t think in metric when I don’t have to, so to convert that to pounds, two kilograms is 4.41 pounds and 103.9 cubic centimeters is 6.34 cubic inches. I guess for the purposes of this discussion the actual shape of the projectile doesn’t matter, but I had always imagined it as a cylinder. A two kilo/4.41 pound cylinder of tungsten would be 28.75mm x 160mm /1.132 in x 6.299 inches. Doesn’t sound very big, does it? Well, that’s just the projectile, next we must impart some velocity on it. Light speed, it the galactic speed limit, nothing can exceed that, at least not under current physics. Light speed is 299,792,458 meters per second or roughly 186,000 miles per second. Woof, fast I know, but we’re not going to push this tungsten quite that fast. Electricity travels at the speed of light, but computers take time to process, and capacitors take time to cycle and coils to energize and then you have to consider the tolerances and variability that comes into any mass-produced engineering project. I estimated that the electromagnetic coils inside a spaceship’s rail gun powered by a nuclear reactor or six would be able to embark a speed of 24% of light speed. No, I didn’t do any math for that, it’s just a gut feeling estimation. For argument’s sake, for the lore of the world, the Russian Federation can build rail guns that can accelerate a 2-kilo slug of tungsten 28.75x160mm to 24% of light speed.

24 percent of light speed is that’s 44,640 miles per second or 71,950,189.92 meters per second. Still really fast. However, that formula I described earlier won’t work. Sure we could convert the kilos and pounds into grains, and the miles per second into feet per second, but that 450440 number is a constant that represents Earth’s atmosphere and gravity at sea level. This rail gun is being fired from one ship to another, without the interference of an atmosphere, and it’s not over Earth either. Fortunately, the ship firing the rail gun is at a lower orbital altitude and is fighting at a second ship at a higher orbital altitude, so the effect of gravity is working in the inverse direction of the projectile. This planet’s gravity is roughly 1.7 that of Earth’s gravity, and we all know Earth’s gravity exerts a force of 9.8 meters per second, so this other planet will exert a force of 16.66 meters per second on the projectile, slowing it down. For reference, this combat is in orbit over the planet Kepler 452B which is estimate to be roughly 1.9 earth masses, but  I figure in order to make space flight and general human occupation more feasible there I would lighten up the gravity and attribute the current ‘over estimation’ of its mass to things like its moon and asteroids in the solar system increasing the solar wobble that was in part used to discover the existence of and calculate the mass of the planet. Anyway… back to giant space guns.

Without factoring atmospheric interference the kinetic energy formula is a simple KE=.5 x m x v^2  where M=mass and V = velocity so in this case, we have .5 the mass of two kilos (one kilogram) multiplied by the velocity of 71,950,189.92 meters per second

KE= 1 kilo x 5,176,829,829,524,069.6064 meters per second but also have to subtract the 16.66 meters per second from Kepler 452B’s gravity. So then the end result is

KE=5,176,829,829,523,792.05 Jules (3,818,233,735,506,251.5 foot pounds) now that’s a big number… that’s roughly 2,089,892,575,537 times more powerful then the deer rifle mentioned in the initial formula/example.

That is a number so big, that it will be operating on a whole different scale. That’s like a meteor hitting a planet level of kinetic energy and I find, at least for the way my head works, it best to compare this to explosives and convert it to TNT-related units like they do with nuclear devices. Using some calculator I found on the internet and no specific formula I am told that 5,176,829,829,523,792 Jules is the same as 1,237,292 tons of TNT or as would be said in colloquial speak “1.2 megatons of TNT”. That’s a big boom the little boy dropped on Hiroshima was 15 kilotons of TNT. This is nearly ten times as powerful as that, but it’s purely kinetic energy, no radiation, no explosion, just wallop.

But I didn’t stop there…. These two space ships in orbit are some distance apart right? If one ship is in a low planetary orbit, and the other is slightly higher, for this instance I chose an altitude difference of 200 kilometers, that’s a safe orbital distance separation so you don’t have to worry about colliding with each other. At the 24% of light speed chosen velocity flight time of the projectile is .00278 seconds. That’s not enough time for the targeted ship to get out of harm’s way. To use a larger example, if fired from the Earth, across the 382500-kilometer gap to the moon, the projectiles flight time would only be .0053 seconds. The same velocity would get you from the earth to Mars in just about thirteen seconds.

I know I could have been more accurate in my calculations since the planet’s gravity would be 16.6 meters per second on the surface, but if the two ships mentioned are orbiting at 500 and 700-kilometer altitudes the gravity will be less up there.

That was the easy part of the problem, for the story’s sake. The next question is what left me hung up for months. If that small of a projectile actually hit the target at that velocity… what would happen to the target? Would the tungsten punch a little hole through the targeted ship and go off into space leaving a 29ish mm hole in the hull, or would it impart some or all of that kinetic energy and convert the spaceship into a cloud of orbital glitter?

I spent months crawling through forums, social media platforms, and so on talking to engineers and people better at math than I in order to try and answer that question. Finally, I met a NASA engineer who was lurking in one of the dark quiet nerdy corners of Discord, and after some discussion, the answer I got was “it depends”. As the author, I took some creative liberties at that point, and you’ll just have to read the third book, when it comes out to see what happened. (*wink wink*)

Thanks for stopping by, I’ll see you out there.

Published by chacerandolph

Science fiction author and Avionics Technician

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: